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\title{实变函数练习5.4 - 一般可测函数的勒贝格积分}
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\date{2024 年 5 月 20 日}
%\date{March 9, 2021}

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\begin{document}

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\begin{enumerate}

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\item  %Problem 01
设 $E\subseteq\mathbb{R}^n$ 是可测集，设 $f(x)$ 是 $E$ 上的可测函数，  
写出勒贝格积分的定义， $$\int_E f(x)dx.$$
什么时候称 $f(x)$ 在 $E$ 上的勒贝格积分是确定的？什么时候称 $f(x)$ 在 $E$ 上是勒贝格可积的？

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%\item  %Problem 02、S5.4定理1
%设 $E\subset\mathbb{R}^n$ 是可测集，则有 
%\begin{enumerate}
%\item   若 $E\neq\varnothing$ 但 $m(E)=0$, 则 $E$ 上的任何实函数 $f$ 都是勒贝格可积的，且积分为零。  
%\item   若 $f\in L(E)$, 则 $f(x)$ 在 $E$ 上是几乎处处有限的。 
%\item   设 $f\in L(E)$, 设 $E=A\cup B$, 其中 $A,B$ 是 $E$ 的两个互不相交的可测集，则 $$\int_E f(x)dx = \int_A f(x)dx + \int_B  f(x)dx. $$  
%\item   设 $f\in L(E)$, 设 $f(x)=g(x)$ 在 $E$ 上几乎处处成立，则 $g\in L(E)$, 且 $$\int_E f(x)dx = \int_E g(x)dx. $$ 
%
%\item   设 $f,g\in L(E)$, 设 $f(x)\le g(x)$ 在 $E$ 上几乎处处成立，则 
%$$\int_E f(x)dx \le \int_E g(x)dx. $$ 
%特别地，设 $m(E)<\infty$, 且 $b\le f(x)\le B$ 在 $E$ 上几乎处处成立，则 
%$$bm(E) \le \int_E f(x)dx \le Bm(E). $$ 
%
%\end{enumerate}
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%\item  %Problem 03、S5.4定理2
%设 $E\subset\mathbb{R}^n$ 是可测集，设 $f,g\in L(E)$, 则有 
%\begin{enumerate}
%\item   对任意实数 $\lambda$, 有 $\lambda f\in L(E)$, 且 $$\int_E \lambda f(x)dx = \lambda \int_E f(x)dx. $$ 
%\item   有 $f+g\in L(E)$, 且 $$\int_E [ f(x) + g(x)]dx = \int_E f(x)dx + \int_E g(x)dx. $$ 
%\item   对任意实数 $\alpha,\beta$, 有 $\alpha f + \beta g\in L(E)$, 且 $$\int_E [\alpha f(x) + \beta g(x)]dx = \alpha\int_E f(x)dx + \beta\int_E g(x)dx. $$ 
%
%\end{enumerate}
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\item  %Problem 04、S5.4定理3
（积分的绝对连续性）设 $E\subseteq\mathbb{R}^n$ 是可测集，设 $f\in L(E)$, 则对任意 $\varepsilon>0$, 存在 $\delta>0$, 使得对任意可测集 $A\subseteq E$, 只要 $m(A)<\delta$, 就有
$$\left\vert \int_A f(x)dx \right\vert \le \int_A \left\vert f(x) \right\vert dx<\varepsilon. $$ 

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\item  %Problem 05、S5.4定理4
（积分的可数可加性）
设有互不相交的可测集 $E_n\subseteq \mathbb{R}^n$,  
设 $f$ 在 $E=\cup_{n=1}^{\infty} E_n$ 上的积分确定，则 
$$ \int_E f(x)dx =\sum\limits_{n=1}^{\infty} \int_{E_n} f(x)  dx. $$ 

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\item  %Problem 06、S5.4定理5
（勒贝格控制收敛定理）设 $E\subseteq \mathbb{R}^n$ 是可测集，设 $\{f_n(x)\}$ 是 $E$ 上的一列可测函数，设 $F(x)$ 是 $E$ 上的非负勒贝格可积函数。设对任意的正整数 $n$, $|f_n(x)|\le F(x)$ 在 $E$ 上几乎处处成立，且 $\lim\limits_{n\to\infty} f_n(x)=f(x)$ 在 $E$ 上几乎处处成立，则
\begin{eqnarray*}
\lim\limits_{n\to\infty} \int_E |f_n(x)-f(x)|dx = 0, \hspace{0.3cm}
\lim\limits_{n\to\infty} \int_E f_n(x) dx = \int_E f(x) dx. 
\end{eqnarray*}


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%\item  %Problem 07、S5.4定理6
%设 $E\subset\mathbb{R}^n$ 是可测集，设 $\{f_n(x)\}$ 与 $f(x)$ 都是 $E$ 上的可测函数，
%设 $F(x)$ 是 $E$ 上的非负勒贝格可积函数。
%设对任意的正整数 $n$, $|f_n(x)|\le F(x)$ 在 $E$ 上几乎处处成立，且 $f_n(x)$ 在 $E$ 上依测度收敛于 $f(x)$, 则
%\begin{eqnarray*}
%\lim\limits_{n\to\infty} \int_E |f_n(x)-f(x)|dx &=& 0, \\ 
%\lim\limits_{n\to\infty} \int_E f_n(x) dx &=& \int_E f(x) dx. 
%\end{eqnarray*}
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%\vspace{0.1cm}

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\item  %Problem 08、S5.4定理7
设 $E\subseteq\mathbb{R}^n$ 是可测集，设 $\{f_n(x)\}$ 是 $E$ 上的一列勒贝格可积函数。
%设正项级数 $$\sum\limits_{n=1}^{\infty} \int_E |f_n(x)|dx$$ 收敛，
%则函数项级数 $$\sum\limits_{n=1}^{\infty} f_n(x)$$ 在 $E$ 上几乎处处收敛，
%其和函数 $F(x)$ 在 $E$ 上勒贝格可积，且 
给出条件使积分与级数交换，
$$ \int_E \left( \sum\limits_{n=1}^{\infty} f_n(x) \right) dx =\sum\limits_{n=1}^{\infty} \int_E f_n(x) dx. $$ 

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\item  %Problem 09、S5.4定理8
设 $E\subseteq\mathbb{R}^n$ 是可测集，设 $f(x,t)$ 是 $E\times (a,b)$ 上的实函数。
%设对任意的 $t\in (a,b)$, 作为 $x$ 的函数 $f(x,t)$ 在 $E$ 上是勒贝格可积的，
%设对任意的 $x\in E$, 作为 $t$ 的函数 $f(x,t)$ 在 $(a,b)$ 上是可导的，且 
%$$\left\vert \frac{\partial }{\partial t}(f(x,t)) \right\vert \le F(x), $$
%这里 $F(x)$ 是 $E$ 上的某个非负勒贝格可积函数。则 $\int_E f(x,t)dx$ 作为 $t$ 的函数在 $(a,b)$ 中可导，且
给出条件使求导与积分交换，
$$\frac{d}{dt} \int_E f(x,t)dx = \int_E \frac{\partial }{\partial t}(f(x,t)) dx. $$

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%\item  %Problem 10、S5.4例子
%设 $f\in L[a,b]$, 则对任意 $\varepsilon>0$, 存在 $g\in C[a,b]$, 使得
%$$\int_{[a,b]} |f(x)-g(x)|dx < \varepsilon. $$
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\item  %Problem 12、S5习题7
设 $f(x) = \frac{\sin(1/x)}{x^\alpha},\,\, 0<x\le 1,$
讨论 $\alpha$ 为何值时，$f(x)$ 在 $(0,1]$ 上是勒贝格可积的。

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\item  %Problem 13、S5习题12
试从 $\frac{1}{1+x} = (1-x) + (x^2-x^3) + \cdots (0<x<1)$ 证明
$\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $. 

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\item  %Problem 14、S5习题13
证明：
$$\lim\limits_{n\to\infty} \int_0^\infty \frac{dt}{\left( 1+\frac{t}{n} \right)^n t^{\frac{1}{n}}} =1. $$

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\item  %Problem 15、S5习题14
若 $p>-1$, 证明：
$$\int_0^1 \frac{x^p}{1-x}\ln\frac{1}{x}dx =\sum\limits_{n=1}^{\infty} \frac{1}{(p+n)^2}. $$

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\end{enumerate}


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\end{document}

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